Find the LCD x/(3x+6) , (x+1)/(2x^2-8)

$\frac{x}{3 x + 6}$ , $\frac{x + 1}{2 x^{2} - 8}$

Step 1

Simplify each polynomial.

Factor $3$ out of $3 x + 6$ .

Factor $3$ out of $3 x$ .

$\frac{x}{3 (x) + 6}, \frac{x + 1}{2 x^{2} - 8}$

Factor $3$ out of $6$ .

$\frac{x}{3 x + 3 \cdot 2}, \frac{x + 1}{2 x^{2} - 8}$

Factor $3$ out of $3 x + 3 \cdot 2$ .

$\frac{x}{3 (x + 2)}, \frac{x + 1}{2 x^{2} - 8}$

Simplify the denominator.

Factor $2$ out of $2 x^{2} - 8$ .

Factor $2$ out of $2 x^{2}$ .

$\frac{x}{3 (x + 2)}, \frac{x + 1}{2 (x^{2}) - 8}$

Factor $2$ out of $- 8$ .

$\frac{x}{3 (x + 2)}, \frac{x + 1}{2 x^{2} + 2 \cdot - 4}$

Factor $2$ out of $2 x^{2} + 2 \cdot - 4$ .

$\frac{x}{3 (x + 2)}, \frac{x + 1}{2 (x^{2} - 4)}$

Rewrite $4$ as $2^{2}$ .

$\frac{x}{3 (x + 2)}, \frac{x + 1}{2 (x^{2} - 2^{2})}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$\frac{x}{3 (x + 2)}, \frac{x + 1}{2 (x + 2) (x - 2)}$

Step 2

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$3 (x + 2), 2 (x + 2) (x - 2)$

Step 3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 4

Since $3$ has no factors besides $1$ and $3$ .

$3$ is a prime number

Step 5

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

Step 6

The LCM of $3, 2$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2 \cdot 3$

Step 7

Multiply $2$ by $3$ .

$6$

Step 8

The factor for $x + 2$ is $x + 2$ itself.

$(x + 2) = x + 2$

$(x + 2)$ occurs $1$ time.

Step 9

The factor for $x - 2$ is $x - 2$ itself.

$(x - 2) = x - 2$

$(x - 2)$ occurs $1$ time.

Step 10

The LCM of $x + 2, x + 2, x - 2$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x + 2) (x - 2)$

Step 11

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$6 (x + 2) (x - 2)$

Find the LCD x/(3x+6) , (x+1)/(2x^2-8)

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